Multiply the following complex numbers: $({4-i}) \cdot ({-4-4i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4-i}) \cdot ({-4-4i}) = $ $ ({4} \cdot {-4}) + ({4} \cdot {-4}i) + ({-1}i \cdot {-4}) + ({-1}i \cdot {-4}i) $ Then simplify the terms: $ (-16) + (-16i) + (4i) + (4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -16 + (-16 + 4)i + 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -16 + (-16 + 4)i - 4 $ The result is simplified: $ (-16 - 4) + (-12i) = -20-12i $